3 Answers
3

Expand $\cos(x-y)$ into $\cos(x)\cos(y)+\sin(x)\sin(y)$ to show that $f(x)=a\cos(x)+b\sin(x)$ with
$$
a=\lambda\int_0^{\pi/2}\cos(y)f(y)dy,\qquad b=\lambda\int_0^{\pi/2}\sin(y)f(y)dy.
$$
Since
$$
\int_0^{\pi/2}\cos^2(y)dy=\int_0^{\pi/2}\sin^2(y)dy=\frac{\pi}4,\qquad
\int_0^{\pi/2}\cos(y)f(y)dy=\frac12,
$$
this yields
$$
4a=\lambda(a\pi+2b),\qquad 4b=\lambda(2a+\pi b).
$$
If $\lambda\pi-4\ne\pm2\lambda$, the only solution is $a=b=0$, hence $f=0$. If $\lambda\pi-4=\pm2\lambda$, then $\lambda\ne0$ hence every $a=\mp b$ is a solution. For $\lambda=4/(\pi\pm2)$, $f(x)=a(\cos x\pm\sin x)$, or, equivalently, $f(x)=c\cos(x\mp\pi/4)$.

A different solution from the one by nbubis is differentiation under the integral sign. Hence we get,
\begin{equation}
\frac{df(x)}{dx} = - \lambda \int_0^{\pi/2} \sin({x-y)}f(y)\,dy, \qquad \lambda \in \mathbb R.
\end{equation}

You can try to expand $f(x)$ as a Fourier series, and then compare the result, term by term. You will then see that the all higher orders $(n > 1)$ must be zero, since the integral only yields terms of order $1$.

So we now know that $f(x)$ must be of the form:
$$f(x) = a \sin(x) + b \cos(x)$$
And inserting this leads to the following equations for $a,b$:
$$ a = \lambda\left(\frac{\pi a}{4} + \frac{b}{2}\right)\\
b = \lambda\left(\frac{\pi b}{4} + \frac{a}{2}\right)$$
Unless $\lambda$ is one of two values, $a,b$ must be zero, and so for almost all values of lambda, the only function satisfying your equation is $f(x) = 0$.